CAIE P3 2020 June — Question 3 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with tan(θ ± α)
DifficultyStandard +0.3 This question requires applying the tan addition formula twice, algebraic manipulation to form a quadratic in tan θ, then solving. While it involves multiple steps and careful algebra, it's a standard application of a core P3 technique with no novel insight required. The 'hence solve' structure guides students through the method, making it slightly easier than average.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

3 Express the equation \(\tan \left( \theta + 60 ^ { \circ } \right) = 2 + \tan \left( 60 ^ { \circ } - \theta \right)\) as a quadratic equation in \(\tan \theta\), and hence solve the equation for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

Question 3:
AnswerMarks Guidance
AnswerMark Guidance
Use \(\tan(A \pm B)\) formula and obtain an equation in \(\tan\theta\)M1
Using \(\tan 60° = \sqrt{3}\), obtain a horizontal equation in \(\tan\theta\) in any correct formA1
Reduce the equation to \(3\tan^2\theta + 4\tan\theta - 1 = 0\), or equivalentA1
Solve a 3-term quadratic for \(\tan\theta\)M1
Obtain a correct answer, e.g. \(12.1°\)A1
Obtain a second correct answer, e.g. \(122.9°\), and no others in the given intervalA1
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Use $\tan(A \pm B)$ formula and obtain an equation in $\tan\theta$ | M1 | |
| Using $\tan 60° = \sqrt{3}$, obtain a horizontal equation in $\tan\theta$ in any correct form | A1 | |
| Reduce the equation to $3\tan^2\theta + 4\tan\theta - 1 = 0$, or equivalent | A1 | |
| Solve a 3-term quadratic for $\tan\theta$ | M1 | |
| Obtain a correct answer, e.g. $12.1°$ | A1 | |
| Obtain a second correct answer, e.g. $122.9°$, and no others in the given interval | A1 | |

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3 Express the equation $\tan \left( \theta + 60 ^ { \circ } \right) = 2 + \tan \left( 60 ^ { \circ } - \theta \right)$ as a quadratic equation in $\tan \theta$, and hence solve the equation for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P3 2020 Q3 [6]}}